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प्रश्न
If \[A = \begin{bmatrix}\alpha & \beta \\ \gamma & - \alpha\end{bmatrix}\] is such that A2 = I, then
पर्याय
1 + α2 + βγ = 0
1 − α2 + βγ = 0
1 − α2 − βγ = 0
1 + α2 − βγ = 0
उत्तर
1 − α2 − βγ = 0
\[Here, \]
\[ A^2 = I\]
\[ \Rightarrow \begin{bmatrix}\alpha & \beta \\ \gamma & - \alpha\end{bmatrix}\begin{bmatrix}\alpha & \beta \\ \gamma & - \alpha\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}\alpha^2 + \beta\gamma & \alpha\beta - \beta\alpha \\ \lambda\alpha - \alpha\gamma & \gamma\beta + \alpha^2\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}\alpha^2 + \beta\gamma & 0 \\ 0 & \gamma\beta + \alpha^2\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[\]
`"The corresponding elements of two equal matrices are equal ."`
\[ \Rightarrow \alpha^2 + \beta\gamma = 1 \]
\[ \Rightarrow 1 - \alpha^2 - \beta\gamma = 0 \]
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